Solution to Several Split Quaternion Matrix Equations

Abstract

Split quaternions have various applications in mathematics, computer graphics, robotics, physics, and so on. In this paper, two useful, real representations of a split quaternion matrix are proposed. Based on this, we derive their fundamental properties. Then, via the real representation method, we obtain the necessary and sufficient conditions for the existence of solutions to two split quaternion matrix equations. In addition, two experimental examples are provided to show their feasibility.

1. Introduction

Split quaternions (or coquaternions) were found in 1849 by James Cockle. The set of all split quaternions over the real number field $\mathbb{R}$ is usually written as

$${\mathbb{H}}_s=\{z_1+z_2\mathbf{i}+z_3\mathbf{j}+z_4\mathbf{k}|{\mathbf{i}}^2=-1,{\mathbf{j}}^2={\mathbf{k}}^2=1,\mathbf{ij}=-\mathbf{ji}=\mathbf{k},z_1,z_2,z_3,z_4\in \mathbb{R}\}.$$

It is well known that ${\mathbb{H}}_s$ is an associative, non-commutative, and four-dimensional Clifford algebra that includes zero divisors, nilpotent elements, and nontrivial idempotents. ${\mathbb{H}}_s$ has various applications in computer graphics [1], robotics [2], physics [3], and so on [4,5,6,7].

The theory of a split quaternion matrix has been studied with a focus on various aspects, such as in the following research: Erdoğdu et al. [8] explored the matrix exponential $e^A$ with a complex adjoint. Guo et al. [9] solved the eigen-problems of $\mathbf{i}$-Hermitian split quaternion matrices due to their important applications on the Schrödinger equation, which is the fundamental equation of quantum mechanics. Jiang et al. [10] studied the right eigenvalues and eigenvectors of split quaternion matrices and derived explicit formulas. Kyrchei [11] provided the original Cramer’s rule for some Hermitian split matrix equations using noncommutative determinants. Wang et al. [12] derived the singular value decomposition of a split quaternion matrix and considered the constraint least squares problem. Solving a split quaternion matrix equation has also received more and more attention. For example, Yue et al. [13] considered the existence of solutions to the split quaternion matrix equation $X-A{X}^{★}B=C$ with $X^{★}\in \{X^{\ast },X^{\mathbf{i}\ast },X^{\mathbf{j}\ast },X^{\mathbf{k}\ast }\}$. Liu et al. [14] discussed the split quaternion matrix equation $AX-X^{★}B=CY+D$ and $X-A{X}^{★}B=CY+D$ with $X^{★}\in \{X,X^{\mathbf{i}},X^{\mathbf{j}},X^{\mathbf{k}}\}$. More equations can be found: $X=\pm X^{★},X^{★}\in \left\{X^{\mathbf{i}\ast },X^{\mathbf{j}\ast },X^{\mathbf{k}\ast }\right\}$ to $(AXB,CXD)=(E,F)$ in Li et al. [15], and $X=X^{\ast }$ solutions to $AXB+CXD=E$ in Yuan et al. [16]. Note that the theory of a split quaternion matrix equation has some background in the area of computer graphics, robotics, or physics. For instance, we can represent a color image as a split quaternion matrix, the matrix equation $HX+N=Y$ can be utilized image restoration task [17], etc. It is important to note that the specific matrix equation form will be adjusted according to the task and algorithm. Thus, motivated by the above works, we aimed to consider the unsolved split quaternion matrix equations:

$$AXB+{\left(AXB\right)}^{★}=E$$

(1)

and

$$X^{★}+CXD=E,$$

(2)

where $★\in \{1,\mathbf{i},\mathbf{j},\mathbf{k},\ast ,\mathbf{i}\ast ,\mathbf{j}\ast ,\mathbf{k}\ast \}$.

This paper is organized as follows: In Section 2, we describe some properties of the real representations of split quaternion matrices. In Section 3 and Section 4, we study the existence and expressions of the solutions to Equations (1) and (2) using the real representations ${(·)}^{\tau }$ and ${(·)}^{\sigma }$ separately. Then, two experimental examples are shown in Section 5, and a conclusion is presented in Section 6.

2. Real Representations

Let ${\mathbb{H}}_s^{m\times n}$ denote the set of split quaternion matrices with the size of $m\times n$. For any $Z\in {\mathbb{H}}_s^{m\times n}$, Z can be represented as $Z=Z_1+Z_2\mathbf{i}+Z_3\mathbf{j}+Z_4\mathbf{k}$, where $Z_1,Z_2,Z_3,Z_4\in {\mathbb{R}}^{m\times n}$. We define three corresponding $\eta$-conjugates as follows:

$$\begin{array}{ccccc}{Z}^{\mathbf{i}}\hfill & =\hfill & {\mathbf{i}}^{-1}Z\mathbf{i}\hfill & =\hfill & Z_1+Z_2\mathbf{i}-Z_3\mathbf{j}-Z_4\mathbf{k},\hfill \\ Z^{\mathbf{j}}\hfill & =\hfill & {\mathbf{j}}^{-1}Z\mathbf{j}\hfill & =\hfill & Z_1-Z_2\mathbf{i}+Z_3\mathbf{j}-Z_4\mathbf{k},\hfill \\ Z^{\mathbf{k}}\hfill & =\hfill & {\mathbf{k}}^{-1}Z\mathbf{k}\hfill & =\hfill & Z_1-Z_2\mathbf{i}-Z_3\mathbf{j}+Z_4\mathbf{k}.\hfill \end{array}$$

Let $Z^{\ast }={Z_1}^T-{Z_2}^T\mathbf{i}-{Z_3}^T\mathbf{j}-{Z_4}^T\mathbf{k}$ be the usual conjugate transpose of Z, where ${(·)}^T$ is the transpose operator of the matrix $(·)$. The other three $\eta$-conjugate transposes of Z are defined as follows:

$$\begin{array}{ccccc}{Z}^{\mathbf{i}\ast }\hfill & =\hfill & {\mathbf{i}}^{-1}{Z}^{\ast }\mathbf{i}& =\hfill & {Z_1}^T-{Z_2}^T\mathbf{i}+{Z_3}^T\mathbf{j}+{Z_4}^T\mathbf{k},\hfill \\ Z^{\mathbf{j}\ast }\hfill & =\hfill & {\mathbf{j}}^{-1}{Z}^{\ast }\mathbf{j}& =\hfill & {Z_1}^T+{Z_2}^T\mathbf{i}-{Z_3}^T\mathbf{j}+{Z_4}^T\mathbf{k},\hfill \\ Z^{\mathbf{k}\ast }\hfill & =\hfill & {\mathbf{k}}^{-1}{Z}^{\ast }\mathbf{k}& =\hfill & {Z_1}^T+{Z_2}^T\mathbf{i}+{Z_3}^T\mathbf{j}-{Z_4}^T\mathbf{k}.\hfill \end{array}$$

To solve our problems in a unified way, we introduce two new real representations of the split quaternion matrices, which are more convenient in solving the eight cases concurrently (see [18]).

Definition 1.

For $Z=Z_1+Z_2\mathbf{i}+Z_3\mathbf{j}+Z_4\mathbf{k}\in {\mathbb{H}}_s^{m\times n}$, we define two real representations of Z as follows:

$$Z^{\tau }=\left[\begin{array}{cccc}{Z}_1& -Z_2& Z_3& Z_4\\ Z_2& Z_1& Z_4& -Z_3\\ Z_3& Z_4& Z_1& -Z_2\\ Z_4& -Z_3& Z_2& Z_1\end{array}\right]\in {\mathbb{R}}^{4m\times 4n},$$

$$Z^{\sigma }=-G_n{Z}^{\tau }=\left[\begin{array}{cccc}{Z}_4& -Z_3& Z_2& Z_1\\ -Z_3& -Z_4& -Z_1& Z_2\\ Z_2& Z_1& Z_4& -Z_3\\ -Z_1& Z_2& -Z_3& -Z_4\end{array}\right]\in {\mathbb{R}}^{4m\times 4n},$$

where

$$G_n=\left[\begin{array}{cccc}0& 0& 0& -I_n\\ 0& 0& I_n& 0\\ 0& -I_n& 0& 0\\ I_n& 0& 0& 0\end{array}\right].$$

The following special matrices will be used to discuss the properties of the above two real representations.

$$R_n=\left[\begin{array}{cccc}0& -I_n& 0& 0\\ I_n& 0& 0& 0\\ 0& 0& 0& I_n\\ 0& 0& -I_n& 0\end{array}\right],Q_n=\left[\begin{array}{cccc}0& 0& I_n& 0\\ 0& 0& 0& I_n\\ I_n& 0& 0& 0\\ 0& I_n& 0& 0\end{array}\right],S_n=\left[\begin{array}{cccc}0& 0& 0& -I_n\\ 0& 0& I_n& 0\\ 0& I_n& 0& 0\\ -I_n& 0& 0& 0\end{array}\right].$$

It is easy to verify that

$$R_n{G}_n=G_n{R}_n^T,Q_n{G}_n=G_n{Q}_n^T,S_n{G}_n=-G_n{S}_n^T,$$

$$G_n{\left(R_n^T\right)}^{-1}=R_n^{-1}{G}_n,G_n{\left(Q_n^T\right)}^{-1}=Q_n^{-1}{G}_n,G_n{\left(S_n^T\right)}^{-1}=-S_n^{-1}{G}_n.$$

Next, we list some useful properties of the real representations ${(·)}^{\tau }$ and ${(·)}^{\sigma }$, which are easily verified.

Proposition 1.

Let $A,B\in {\mathbb{H}}_s^{m\times n},C\in {\mathbb{H}}_s^{n\times p},a\in \mathbb{R}$. Then,

(a) 

(i)    ${(A+B)}^{\tau }=A^{\tau }+B^{\tau },{\left(AC\right)}^{\tau }=A^{\tau }{C}^{\tau },{\left(aA\right)}^{\tau }=a{A}^{\tau };$

(ii)   ${(A+B)}^{\sigma }=A^{\sigma }+B^{\sigma },{\left(AC\right)}^{\sigma }=A^{\sigma }{G}_n{C}^{\sigma },{\left(aA\right)}^{\sigma }=a{A}^{\sigma };$

(b) 

(i)    ${R_m}^{-1}{A}^{\tau }{R}_n=A^{\tau },{Q_m}^{-1}{A}^{\tau }{Q}_n=A^{\tau },{S_m}^{-1}{A}^{\tau }{S}_n=A^{\tau };$

(ii)   ${\left({R_m}^T\right)}^{-1}{A}^{\sigma }{R}_n=A^{\sigma },{\left({Q_m}^T\right)}^{-1}{A}^{\sigma }{Q}_n=A^{\sigma },{\left({S_m}^T\right)}^{-1}{A}^{\sigma }{S}_n=-A^{\sigma };$

(c) 

${\left(A^{\ast }\right)}^{\tau }=-G_n{\left(A^{\tau }\right)}^T{G}_m,{\left(A^{\ast }\right)}^{\sigma }=-{\left(A^{\sigma }\right)}^T.$

(d) 

Let $Z\left({\mathbb{H}}_s\right)$ be the set of zero divisors of ${\mathbb{H}}_s$, and $A\in Z{\mathbb{H}}_s^{m\times n},C\in Z{\mathbb{H}}_s^{n\times p}$, such that $AC=0.$ Then, ${\left(AC\right)}^{\tau }=0$ and ${\left(AC\right)}^{\sigma }=0.$

Proof. 

We only prove the following (d): Assume that $AC=0$. According to (i) of (a), we have $A^{\tau }{C}^{\tau }={\left(AC\right)}^{\tau }=0.$ Next, we prove $A^{\sigma }{C}^{\sigma }=0.$ Via (ii) of (a) and the property $Z^{\tau }=G_n{Z}^{\sigma }$, we get

$$G_m{\left(AC\right)}^{\sigma }=G_m{A}^{\sigma }{G}_n{C}^{\sigma }=A^{\tau }{C}^{\tau }=0.$$

Note that $G_m$ is nonsingular, so we have

$${\left(AC\right)}^{\sigma }=0.$$

□

In the remaining part of the paper, we will see that ${(·)}^{\sigma }$ is suitable for solving the split quaternion matrix equation with the case of $★=\eta$, while ${(·)}^{\tau }$ is more suitable for solving the split quaternion matrix equation with the case of $★=\eta \ast$.

3. Split Quaternion Matrix Equation $\mathit{AXB}+{\left(\mathit{AXB}\right)}^{★}=\mathit{E}$

In this section, we consider the split quaternion matrix Equation (1) in the case of $★\in \{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$ using the real representation ${(·)}^{\sigma }$.

Theorem 1.

Let $A,B,E\in {\mathbb{H}}_s^{n\times n}$ and $★=\eta ,\eta \in \{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$. The split quaternion matrix Equation (1) has a solution $X\in {\mathbb{H}}_s^{n\times n}$ if and only if its corresponding real matrix equation

$$\mathcal{AYB}+\mathcal{CYD}=\mathcal{E}$$

(3)

has a solution $\mathcal{Y}\in {\mathbb{R}}^{4n\times 4n},$ where $\mathcal{A}=A^{\sigma }{G}_n$, $\mathcal{B}=G_n{B}^{\sigma }$, $\mathcal{C}={\left({\eta }^{-1}A\right)}^{\sigma }{G}_n,\mathcal{D}=G_n{\left(B\eta \right)}^{\sigma }$ and $\mathcal{E}=E^{\sigma }.$

Moreover, if $\mathcal{Y}={\left(Y_{ij}\right)}_{4\times 4}\in {\mathbb{R}}^{4n\times 4n}$ is a solution to the real matrix Equation (3), then

$$\begin{array}{c}\hfill X=\frac{1}{4}[(Y_{14}-Y_{23}+Y_{32}-Y_{41})+(Y_{13}+Y_{24}+Y_{31}+Y_{42})\mathbf{i}\\ \hfill +(-Y_{12}-Y_{21}-Y_{34}-Y_{43})\mathbf{j}+(Y_{11}-Y_{22}+Y_{33}-Y_{44})\mathbf{k}],\end{array}$$

or, equivalently,

$$\begin{array}{c}\hfill X=\frac{1}{16}\left[\begin{array}{cccc}{I}_n\mathbf{k}& -I_n\mathbf{j}& I_n\mathbf{i}& -I_n\end{array}\right](\mathcal{Y}+{R_n}^T\mathcal{Y}{R_n}^{-1}+{Q_n}^T\mathcal{Y}{Q_n}^{-1}-{S_n}^T\mathcal{Y}{S_n}^{-1})\left[\begin{array}{c}{I}_n\\ -I_n\mathbf{i}\\ I_n\mathbf{j}\\ I_n\mathbf{k}\end{array}\right]\end{array}$$

is the solution of (1).

Proof. 

Suppose the split quaternion matrix Equation (1) has a solution, $X\in {\mathbb{H}}_s^{n\times n}$; using ${\left(AXB\right)}^{\eta }={\eta }^{-1}\left(AXB\right)\eta$, it becomes

$$AXB+{\eta }^{-1}AXB\eta =E.$$

(4)

Applying (a) of Proposition 1 to (4) yields

$$A^{\sigma }{G}_n{X}^{\sigma }{G}_n{B}^{\sigma }+{\left({\eta }^{-1}A\right)}^{\sigma }{G}_n{X}^{\sigma }{G}_n{\left(B\eta \right)}^{\sigma }=E^{\sigma },$$

Thus, $X^{\sigma }$ is a solution to (3), i.e., if (1) is solvable, then (3) is also solvable.

Conversely, suppose the real matrix Equation (3) has a solution, $\mathcal{Y}={\left(Y_{ij}\right)}_{4\times 4},Y_{ij}\in {\mathbb{R}}^{n\times n},i,j=1,2,3,4.$ Then, it follows from (b) of Proposition 1 that

$$\begin{array}{cc}\hfill \left({\left(R_n^T\right)}^{-1}{A}^{\sigma }{R}_n\right)G_n\mathcal{Y}{G}_n\left({\left(R_n^T\right)}^{-1}{B}^{\sigma }{R}_n\right)& +\left({\left(R_n^T\right)}^{-1}{\left({\eta }^{-1}A\right)}^{\sigma }{R}_n\right)G_n\mathcal{Y}{G}_n\left({\left(R_n^T\right)}^{-1}{\left(B\eta \right)}^{\sigma }{R}_n\right)\hfill \\ & =\left({\left(R_n^T\right)}^{-1}{E}^{\sigma }{R}_n\right),\hfill \\ \hfill \left({\left(Q_n^T\right)}^{-1}{A}^{\sigma }{Q}_n\right)G_n\mathcal{Y}{G}_n\left({\left(Q_n^T\right)}^{-1}{B}^{\sigma }{Q}_n\right)& +\left({\left(Q_n^T\right)}^{-1}{\left({\eta }^{-1}A\right)}^{\sigma }{Q}_n\right)G_n\mathcal{Y}{G}_n\left({\left(Q_n^T\right)}^{-1}{\left(B\eta \right)}^{\sigma }{Q}_n\right)\hfill \\ & =\left({\left(Q_n^T\right)}^{-1}{E}^{\sigma }{Q}_n\right),\hfill \\ \hfill \left({-\left(S_n^T\right)}^{-1}{A}^{\sigma }{S}_n\right)G_n\mathcal{Y}{G}_n\left({-\left(S_n^T\right)}^{-1}{B}^{\sigma }{S}_n\right)& +(-{\left(S_n^T\right)}^{-1}{\left({\eta }^{-1}A\right)}^{\sigma }{S}_n)G_n\mathcal{Y}{G}_n\left({-\left(S_n^T\right)}^{-1}{\left(B\eta \right)}^{\sigma }{S}_n\right)\hfill \\ & =(-{\left(S_n^T\right)}^{-1}{E}^{\sigma }{S}_n).\hfill \end{array}$$

After canceling ${\left(R_n^T\right)}^{-1},{\left(Q_n^T\right)}^{-1},{\left(S_n^T\right)}^{-1}$ from the left side and $R_n,Q_n,S_n$ from the right side of the above equations and then using

$$R_n{G}_n=G_n{R}_n^T,Q_n{G}_n=G_n{Q}_n^T,S_n{G}_n=-G_n{S}_n^T,$$

$$G_n{\left(R_n^T\right)}^{-1}=R_n^{-1}{G}_n,G_n{\left(Q_n^T\right)}^{-1}=Q_n^{-1}{G}_n,G_n{\left(S_n^T\right)}^{-1}=-S_n^{-1}{G}_n,$$

we have

$$\begin{array}{cc}\hfill A^{\sigma }{G}_n{R_n}^T\mathcal{Y}{R_n}^{-1}{G}_n{B}^{\sigma }+{\left({\eta }^{-1}A\right)}^{\sigma }{G}_n{R_n}^T\mathcal{Y}{R_n}^{-1}{G}_n{\left(B\eta \right)}^{\sigma }& =E^{\sigma },\hfill \\ \hfill A^{\sigma }{G}_n{Q_n}^T\mathcal{Y}{Q_n}^{-1}{G}_n{B}^{\sigma }+{\left({\eta }^{-1}A\right)}^{\sigma }{G}_n{Q_n}^T\mathcal{Y}{Q_n}^{-1}{G}_n{\left(B\eta \right)}^{\sigma }& =E^{\sigma },\hfill \\ \hfill A^{\sigma }{G}_n{S_n}^T\mathcal{Y}{S_n}^{-1}{G}_n{B}^{\sigma }+{\left({\eta }^{-1}A\right)}^{\sigma }{G}_n{S_n}^T\mathcal{Y}{S_n}^{-1}{G}_n{\left(B\eta \right)}^{\sigma }& =-E^{\sigma }.\hfill \end{array}$$

Thus, ${R_n}^T\mathcal{Y}{R_n}^{-1},{Q_n}^T\mathcal{Y}{Q_n}^{-1}$ and $-{S_n}^T\mathcal{Y}{S_n}^{-1}$ are also solutions to (3), and so is

$$\widehat{\mathcal{Y}}:=\frac{1}{4}(\mathcal{Y}+{R_n}^T\mathcal{Y}{R_n}^{-1}+{Q_n}^T\mathcal{Y}{Q_n}^{-1}-{S_n}^T\mathcal{Y}{S_n}^{-1}).$$

Furthermore, we can calculate $\widehat{\mathcal{Y}}$ in terms of $\left\{Y_{ij}\right\}$ and set

$$\widehat{\mathcal{Y}}:=\left[\begin{array}{cccc}{Z}_4& -Z_3& Z_2& Z_1\\ -Z_3& -Z_4& -Z_1& Z_2\\ Z_2& Z_1& Z_4& -Z_3\\ -Z_1& Z_2& -Z_3& -Z_4\end{array}\right],$$

where

$$Z_1=\frac{1}{4}(Y_{14}-Y_{23}+Y_{32}-Y_{41}),Z_2=\frac{1}{4}(Y_{13}+Y_{24}+Y_{31}+Y_{42}),$$

$$Z_3=\frac{1}{4}(-Y_{12}-Y_{21}-Y_{34}-Y_{43}),Z_4=\frac{1}{4}(Y_{11}-Y_{22}+Y_{33}-Y_{44}).$$

Now, we construct a split quaternion matrix, X, using the solution $\widehat{\mathcal{Y}}$:

$$X=Z_1+Z_2\mathbf{i}+Z_3\mathbf{j}+Z_4\mathbf{k}=\frac{1}{4}\left[\begin{array}{cccc}{I}_n\mathbf{k}& -I_n\mathbf{j}& I_n\mathbf{i}& -I_n\end{array}\right]\widehat{\mathcal{Y}}\left[\begin{array}{c}{I}_n\\ -I_n\mathbf{i}\\ I_n\mathbf{j}\\ I_n\mathbf{k}\end{array}\right].$$

(5)

It is easy to verify that $X^{\sigma }=\widehat{\mathcal{Y}},$ i.e., the constructed X satisfies

$$A^{\sigma }{G}_n{X}^{\sigma }{G}_n{B}^{\sigma }+{\left({\eta }^{-1}A\right)}^{\sigma }{G}_n{X}^{\sigma }{G}_n{\left(B\eta \right)}^{\sigma }=E^{\sigma }.$$

Hence, using the inverse of ${(·)}^{\sigma }$, we have $AXB+{\left(AXB\right)}^{\eta }=E$; that is, X in (5) is the solution to (1). Therefore, that (3) is solvable implies that (1) is solvable, too.

In conclusion, we have shown that (1) is solvable if and only if (3) is solvable. Additionally, we have proven that the solutions to (1) can be constructed from the solution $\mathcal{Y}$ of (3) when (3) is solvable. □

Next, we use the real representation of ${(·)}^{\tau }$ to study the split quaternion matrix Equation (1) in the case of $★\in \{\ast ,\mathbf{i}\ast ,\mathbf{j}\ast ,\mathbf{k}\ast \}$.

Theorem 2.

Let $A,B,E\in {\mathbb{H}}_s^{n\times n}$ and $★=\eta \ast ,\eta \in \{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$. The split quaternion matrix Equation (1) has a solution, $X\in {\mathbb{H}}_s^{n\times n}$, if and only if its corresponding real matrix equation,

$$\mathcal{AYB}-\mathcal{C}{\mathcal{Y}}^T\mathcal{D}=\mathcal{E}$$

(6)

has a solution, $\mathcal{Y}\in {\mathbb{R}}^{4n\times 4n}$, where $\mathcal{A}=A^{\tau }$, $\mathcal{B}=B^{\tau }$, $\mathcal{C}={\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }{G}_n,\mathcal{D}=G_n{\left(A^{\ast }\eta \right)}^{\tau }$, and $\mathcal{E}=E^{\tau }.$

Moreover, if $\mathcal{Y}={\left(Y_{ij}\right)}_{4\times 4}\in {\mathbb{R}}^{4n\times 4n}$ is a solution to the real matrix Equation (6), then

$$\begin{array}{c}\hfill X=\frac{1}{4}[(Y_{11}+Y_{22}+Y_{33}+Y_{44})+(-Y_{12}+Y_{21}-Y_{34}+Y_{43})\mathbf{i}\\ \hfill +(Y_{13}-Y_{24}+Y_{31}-Y_{42})\mathbf{j}+(Y_{14}+Y_{23}+Y_{32}+Y_{41})\mathbf{k}],\end{array}$$

(7)

or, equivalently,

$$X=\frac{1}{16}\left[\begin{array}{cccc}{I}_n& I_n\mathbf{i}& I_n\mathbf{j}& I_n\mathbf{k}\end{array}\right](\mathcal{Y}+R_n\mathcal{Y}{R_n}^{-1}+Q_n\mathcal{Y}{Q_n}^{-1}+S_n\mathcal{Y}{S_n}^{-1})\left[\begin{array}{c}{I}_n\\ -I_n\mathbf{i}\\ I_n\mathbf{j}\\ I_n\mathbf{k}\end{array}\right].$$

is the solution to (1).

Proof. 

Suppose the split quaternion matrix Equation (1) has a solution, $X\in {\mathbb{H}}_s^{n\times n}$, after the rewritten,

$$AXB+{\eta }^{-1}{B}^{\ast }{X}^{\ast }{A}^{\ast }\eta =E.$$

(8)

Applying (a) and (c) of Proposition 1 to (8) yields

$$A^{\tau }{X}^{\tau }{B}^{\tau }+{\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }(-G_n{\left(X^{\tau }\right)}^T{G}_n){\left(A^{\ast }\eta \right)}^{\tau }=E^{\tau },$$

or

$$A^{\tau }{X}^{\tau }{B}^{\tau }-\left({\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }{G}_n\right){\left(X^{\tau }\right)}^T\left(G_n{\left(A^{\ast }\eta \right)}^{\tau }\right)=E^{\tau }.$$

Thus, $X^{\tau }$ is a solution to (6). Therefore, if (1) is solvable, then (6) is also solvable.

Conversely, suppose the real matrix Equation (6) has a solution, $\mathcal{Y}=\left(Y_{ij}\right),Y_{ij}\in {\mathbb{R}}^{n\times n},i,j=1,2,3,4.$ Then, it follows from (b) of Proposition 1 that

$$\begin{array}{cc}\hfill \left(R_n^{-1}{A}^{\tau }{R}_n\right)\mathcal{Y}\left(R_n^{-1}{B}^{\tau }{R}_n\right)-\left(R_n^{-1}{\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }{R}_n\right)G_n{\mathcal{Y}}^T{G}_n\left(R_n^{-1}{\left(A^{\ast }\eta \right)}^{\tau }{R}_n\right)& =\left(R_n^{-1}{E}^{\tau }{R}_n\right),\hfill \\ \hfill \left(Q_n^{-1}{A}^{\tau }{Q}_n\right)\mathcal{Y}\left(Q_n^{-1}{B}^{\tau }{Q}_n\right)-\left(Q_n^{-1}{\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }{Q}_n\right)G_n{\mathcal{Y}}^T{G}_n\left(Q_n^{-1}{\left(A^{\ast }\eta \right)}^{\tau }{Q}_n\right)& =\left(Q_n^{-1}{E}^{\tau }{Q}_n\right),\hfill \\ \hfill \left(S_n^{-1}{A}^{\tau }{S}_n\right)\mathcal{Y}\left(S_n^{-1}{B}^{\tau }{S}_n\right))-\left(S_n^{-1}{\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }{S}_n\right)G_n{\mathcal{Y}}^T{G}_n\left(S_n^{-1}{\left(A^{\ast }\eta \right)}^{\tau }{S}_n\right)& =\left(S_n^{-1}{E}^{\tau }{S}_n\right).\hfill \end{array}$$

Next, using

$$R_n{G}_n=-G_n{R}_n,Q_n{G}_n=G_n{Q}_n,S_n{G}_n=-G_n{S}_n,$$

$$G_n{R_n}^{-1}=-{R_n}^{-1}{G}_n,G_n{Q_n}^{-1}={Q_n}^{-1}{G}_n,G_n{S_n}^{-1}=-{S_n}^{-1}{G}_n$$

gives

$$\begin{array}{cc}\hfill \left(R_n^{-1}{A}^{\tau }{R}_n\right)\mathcal{Y}\left(R_n^{-1}{B}^{\tau }{R}_n\right)-R_n^{-1}{\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }{G}_n\left(R_n{\mathcal{Y}}^T{R_n}^{-1}\right)G_n{\left(A^{\ast }\eta \right)}^{\tau }{R}_n& =\left(R_n^{-1}{E}^{\tau }{R}_n\right),\hfill \\ \hfill \left(Q_n^{-1}{A}^{\tau }{Q}_n\right)\mathcal{Y}\left(Q_n^{-1}{B}^{\tau }{Q}_n\right)-Q_n^{-1}{\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }{G}_n\left(Q_n{\mathcal{Y}}^T{Q_n}^{-1}\right)G_n{\left(A^{\ast }\eta \right)}^{\tau }{Q}_n& =\left(Q_n^{-1}{E}^{\tau }{Q}_n\right),\hfill \\ \hfill \left(S_n^{-1}{A}^{\tau }{S}_n\right)\mathcal{Y}\left(S_n^{-1}{B}^{\tau }{S}_n\right))-S_n^{-1}{\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }{G}_n\left(S_n{\mathcal{Y}}^T{S_n}^{-1}\right)G_n{\left(A^{\ast }\eta \right)}^{\tau }{S}_n& =\left(S_n^{-1}{E}^{\tau }{S}_n\right).\hfill \end{array}$$

Premultiplying $R_n,Q_n,S_n$ and postmultiplying ${R_n}^{-1},{Q_n}^{-1},{S_n}^{-1}$, we obtain

$$\begin{array}{cc}\hfill A^{\tau }{R}_n\mathcal{Y}{R_n}^{-1}{B}^{\tau }-{\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }{G}_n\left(R_n{\mathcal{Y}}^T{R_n}^{-1}\right)G_n{\left(A^{\ast }\eta \right)}^{\tau }& =E^{\tau },\hfill \\ \hfill A^{\tau }{Q}_n\mathcal{Y}{Q_n}^{-1}{B}^{\tau }-{\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }{G}_n\left(Q_n{\mathcal{Y}}^T{Q_n}^{-1}\right)G_n{\left(A^{\ast }\eta \right)}^{\tau }& =E^{\tau },\hfill \\ \hfill A^{\tau }{S}_n\mathcal{Y}{S_n}^{-1}{B}^{\tau }-{\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }{G}_n\left(S_n{\mathcal{Y}}^T{S_n}^{-1}\right)G_n{\left(A^{\ast }\eta \right)}^{\tau }& =E^{\tau }.\hfill \end{array}$$

Thus, $R_n\mathcal{Y}{R_n}^{-1},Q_n\mathcal{Y}{Q_n}^{-1}$ and $S_n\mathcal{Y}{S_n}^{-1}$ are also solutions to (6), and so is

$$\widehat{\mathcal{Y}}:=\frac{1}{4}(\mathcal{Y}+R_n\mathcal{Y}{R_n}^{-1}+Q_n\mathcal{Y}{Q_n}^{-1}+S_n\mathcal{Y}{S_n}^{-1}).$$

After simplification, we can get $\widehat{\mathcal{Y}}$ in terms of $\left\{Y_{ij}\right\}$ and set

$$\widehat{\mathcal{Y}}:=\left[\begin{array}{cccc}{Z}_1& -Z_2& Z_3& Z_4\\ Z_2& Z_1& Z_4& -Z_3\\ Z_3& Z_4& Z_1& -Z_2\\ Z_4& -Z_3& Z_2& Z_1\end{array}\right],$$

where

$$Z_1=\frac{1}{4}(Y_{11}+Y_{22}+Y_{33}+Y_{44}),Z_2=\frac{1}{4}(-Y_{12}+Y_{21}-Y_{34}+Y_{43}),$$

$$Z_3=\frac{1}{4}(Y_{13}-Y_{24}+Y_{31}-Y_{42}),Z_4=\frac{1}{4}(Y_{14}+Y_{23}+Y_{32}+Y_{41}).$$

Now, we construct a split quaternion matrix, X, from the solution $\widehat{\mathcal{Y}}$:

$$X=Z_1+Z_2\mathbf{i}+Z_3\mathbf{j}+Z_4\mathbf{k}=\frac{1}{4}\left[\begin{array}{cccc}{I}_n& I_n\mathbf{i}& I_n\mathbf{j}& I_n\mathbf{k}\end{array}\right]\widehat{\mathcal{Y}}\left[\begin{array}{c}{I}_n\\ -I_n\mathbf{i}\\ I_n\mathbf{j}\\ I_n\mathbf{k}\end{array}\right].$$

We can verify that $\widehat{\mathcal{Y}}=X^{\tau }.$ Thus, $X^{\tau }$ satisfies (6), i.e.,

$$A^{\tau }{X}^{\tau }{B}^{\tau }-\left({\left({\eta }^{-1}{B}^{\ast }\right)}^{\tau }{G}_n\right){\left(X^{\tau }\right)}^T\left(G_n{\left(A^{\ast }\eta \right)}^{\tau }\right)=E^{\tau }.$$

Applying the inverse of ${(·)}^{\tau }$ and (c) of Proposition 1, we have

$$AXB+{\eta }^{-1}{B}^{\ast }{X}^{\ast }{A}^{\ast }\eta =E,$$

or

$$AXB+{\left(AXB\right)}^{\eta \ast }=E.$$

Hence, X is a solution to (1). So, that (6) is consistent implies that (1) is also consistent.

In summary, we have shown that (1) is consistent if and only if (6) is consistent. In addition, we have proven that (1) can be constructed from the solution $\mathcal{Y}$ of (6) when (6) is consistent. □

Remark 1.

When $B=I_n$, Theorems 1 and 2 provide the results of the Sylvester equation

$$AX+{\left(AX\right)}^{★}=E,$$

where $★\in \{1,\mathbf{i},\mathbf{j},\mathbf{k},\ast ,\mathbf{i}\ast ,\mathbf{j}\ast ,\mathbf{k}\ast \}$. In particular, when $★=1$, Theorem 1 can be reduced to the equation $AXB=C.$

4. Split Quaternion Matrix Equation ${\mathit{X}}^{★}+\mathit{CXD}=\mathit{E}$

In this section, we consider the split quaternion matrix equation

$$X^{★}+CXD=E$$

in case of $X^{★}\in \{X,X^{\mathbf{i}},X^{\mathbf{j}},X^{\mathbf{k}},X^{\ast },X^{\mathbf{i}\ast },X^{\mathbf{j}\ast },X^{\mathbf{k}\ast }\}$.

Theorem 3.

Let $C,D,E\in {\mathbb{H}}_s^{n\times n}$; then, the split quaternion matrix Equation (2) has a solution, $X\in {\mathbb{H}}_s^{n\times n}$, if and only if the following statements hold:

(a) 

In the case of $X^{★}=X^{\eta },\eta \in \{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$, its corresponding real matrix equation

$$\mathcal{AYB}+\mathcal{CYD}=\mathcal{E}$$

(9)

has a solution, $\mathcal{Y}\in {\mathbb{R}}^{4n\times 4n}$, where $\mathcal{A}={\left({\eta }^{-1}{I}_n\right)}^{\sigma }{G}_n$, $\mathcal{B}=G_n{\left(\eta I_n\right)}^{\sigma }$, $\mathcal{C}=C^{\sigma }{G}_n,\mathcal{D}=G_n{D}^{\sigma }$ and $\mathcal{E}=E^{\sigma }.$ Moreover, if $\mathcal{Y}={\left(Y_{ij}\right)}_{4\times 4}\in {\mathbb{R}}^{4n\times 4n}$ is a solution to the real matrix Equation (9), then

$$\begin{array}{c}\hfill X=\frac{1}{4}[(Y_{14}-Y_{23}+Y_{32}-Y_{41})+(Y_{13}+Y_{24}+Y_{31}+Y_{42})\mathbf{i}\\ \hfill +(-Y_{12}-Y_{21}-Y_{34}-Y_{43})\mathbf{j}+(Y_{11}-Y_{22}+Y_{33}-Y_{44})\mathbf{k}],\end{array}$$

(10)

or, equivalently,

$$X=\frac{1}{16}\left[\begin{array}{cccc}{I}_n\mathbf{k}& -I_n\mathbf{j}& I_n\mathbf{i}& -I_n\end{array}\right](\mathcal{Y}+{R_n}^T\mathcal{Y}{R_n}^{-1}+{Q_n}^T\mathcal{Y}{Q_n}^{-1}-{S_n}^T\mathcal{Y}{S_n}^{-1})\left[\begin{array}{c}{I}_n\\ -I_n\mathbf{i}\\ I_n\mathbf{j}\\ I_n\mathbf{k}\end{array}\right]$$

is the solution of (2).

(b) 

In the case of $X^{★}=X^{\eta \ast },\eta \in \{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$, the split quaternion matrix Equation (2) has a solution, $X\in {\mathbb{H}}_s^{n\times n}$, if and only if its corresponding real matrix equation,

$$\mathcal{Y}-\mathcal{C}{\mathcal{Y}}^T\mathcal{D}=\mathcal{E}$$

(11)

has a solution, $\mathcal{Y}\in {\mathbb{R}}^{4n\times 4n}$, where $\mathcal{C}={\left(D^{\eta \ast }{\eta }^{-1}\right)}^{\tau }{G}_n,\mathcal{D}=G_n{\left(\eta C^{\eta \ast }\right)}^{\tau }$ and $\mathcal{E}={\left(E^{\eta \ast }\right)}^{\tau }.$ Moreover, if $\mathcal{Y}={\left(Y_{ij}\right)}_{4\times 4}\in {\mathbb{R}}^{4n\times 4n}$ is a solution to the real matrix Equation (11), then

$$\begin{array}{c}\hfill X=\frac{1}{4}[(Y_{11}+Y_{22}+Y_{33}+Y_{44})+(-Y_{12}+Y_{21}-Y_{34}+Y_{43})\mathbf{i}\\ \hfill +(Y_{13}-Y_{24}+Y_{31}-Y_{42})\mathbf{j}+(Y_{14}+Y_{23}+Y_{32}+Y_{41})\mathbf{k}],\end{array}$$

or, equivalently,

$$X=\frac{1}{16}\left[\begin{array}{cccc}{I}_n& I_n\mathbf{i}& I_n\mathbf{j}& I_n\mathbf{k}\end{array}\right](\mathcal{Y}+R_n\mathcal{Y}{R_n}^{-1}+Q_n\mathcal{Y}{Q_n}^{-1}+S_n\mathcal{Y}{S_n}^{-1})\left[\begin{array}{c}{I}_n\\ -I_n\mathbf{i}\\ I_n\mathbf{j}\\ I_n\mathbf{k}\end{array}\right]$$

is the solution to (2).

Proof. 

To prove (a), suppose that the split quaternion matrix Equation (2) has a solution, $X\in {\mathbb{H}}_s^{n\times n}$. Using $X^{\eta }={\eta }^{-1}X\eta$, it becomes

$$\left({\eta }^{-1}{I}_n\right)X\left(\eta I_n\right)+CXD=E.$$

(12)

Applying (a) of Proposition 1 to (12) yields

$${\left({\eta }^{-1}{I}_n\right)}^{\sigma }{G}_n{X}^{\sigma }{G}_n{\left(\eta I_n\right)}^{\sigma }+C^{\sigma }{G}_n{X}^{\sigma }{G}_n{D}^{\sigma }=E^{\sigma }.$$

Thus, $X^{\sigma }$ is a solution to (9).

Conversely, suppose that the real matrix Equation (9) has a solution, $\mathcal{Y}={\left(Y_{ij}\right)}_{4\times 4},Y_{ij}\in {\mathbb{R}}^{n\times n},i,j=1,2,3,4.$ Then, it follows from (b) of Proposition 1 that

$$\begin{array}{cc}\hfill \left({\left(R_n^T\right)}^{-1}{\left({\eta }^{-1}{I}_n\right)}^{\sigma }{R}_n\right)G_n\mathcal{Y}{G}_n\left({\left(R_n^T\right)}^{-1}{\left(\eta I_n\right)}^{\sigma }{R}_n\right)& +\left({\left(R_n^T\right)}^{-1}{C}^{\sigma }{R}_n\right)G_n\mathcal{Y}{G}_n\left({\left(R_n^T\right)}^{-1}{D}^{\sigma }{R}_n\right)\hfill \\ & =\left({\left(R_n^T\right)}^{-1}{E}^{\sigma }{R}_n\right),\hfill \\ \hfill \left({\left(Q_n^T\right)}^{-1}{\left({\eta }^{-1}{I}_n\right)}^{\sigma }{Q}_n\right)G_n\mathcal{Y}{G}_n\left({\left(Q_n^T\right)}^{-1}{\left(\eta I_n\right)}^{\sigma }{Q}_n\right)& +\left({\left(Q_n^T\right)}^{-1}{C}^{\sigma }{Q}_n\right)G_n\mathcal{Y}{G}_n\left({\left(Q_n^T\right)}^{-1}{D}^{\sigma }{Q}_n\right)\hfill \\ & =\left({\left(Q_n^T\right)}^{-1}{E}^{\sigma }{Q}_n\right),\hfill \\ \hfill \left({-\left(S_n^T\right)}^{-1}{\left({\eta }^{-1}{I}_n\right)}^{\sigma }{S}_n\right)G_n\mathcal{Y}{G}_n\left({-\left(S_n^T\right)}^{-1}{\left(\eta I_n\right)}^{\sigma }{S}_n\right)& +(-{\left(S_n^T\right)}^{-1}{C}^{\sigma }{S}_n)G_n\mathcal{Y}{G}_n\left({-\left(S_n^T\right)}^{-1}{D}^{\sigma }{S}_n\right)\hfill \\ & =(-{\left(S_n^T\right)}^{-1}{E}^{\sigma }{S}_n).\hfill \end{array}$$

After canceling ${\left(R_n^T\right)}^{-1},{\left(Q_n^T\right)}^{-1},{\left(S_n^T\right)}^{-1}$ from the left side and $R_n,Q_n,S_n$ from the right side of the above equations, and then using

$$R_n{G}_n=G_n{R}_n^T,Q_n{G}_n=G_n{Q}_n^T,S_n{G}_n=-G_n{S}_n^T,$$

$$G_n{\left(R_n^T\right)}^{-1}=R_n^{-1}{G}_n,G_n{\left(Q_n^T\right)}^{-1}=Q_n^{-1}{G}_n,G_n{\left(S_n^T\right)}^{-1}=-S_n^{-1}{G}_n,$$

we obtain

$$\begin{array}{cc}\hfill {\left({\eta }^{-1}{I}_n\right)}^{\sigma }{G}_n{R_n}^T\mathcal{Y}{R_n}^{-1}{G}_n{\left(\eta I_n\right)}^{\sigma }+C^{\sigma }{G}_n{R_n}^T\mathcal{Y}{R_n}^{-1}{G}_n{D}^{\sigma }& =E^{\sigma },\hfill \\ \hfill {\left({\eta }^{-1}{I}_n\right)}^{\sigma }{G}_n{Q_n}^T\mathcal{Y}{Q_n}^{-1}{G}_n{\left(\eta I_n\right)}^{\sigma }+C^{\sigma }{G}_n{Q_n}^T\mathcal{Y}{Q_n}^{-1}{G}_n{D}^{\sigma }& =E^{\sigma },\hfill \\ \hfill {\left({\eta }^{-1}{I}_n\right)}^{\sigma }{G}_n{S_n}^T\mathcal{Y}{S_n}^{-1}{G}_n{\left(\eta I_n\right)}^{\sigma }+C^{\sigma }{G}_n{S_n}^T\mathcal{Y}{S_n}^{-1}{G}_n{D}^{\sigma }& =-E^{\sigma }.\hfill \end{array}$$

We can see that ${R_n}^T\mathcal{Y}{R_n}^{-1},{Q_n}^T\mathcal{Y}{Q_n}^{-1}$ and $-{S_n}^T\mathcal{Y}{S_n}^{-1}$ are also solutions to (9), and so is

$$\widehat{\mathcal{Y}}:=\frac{1}{4}(\mathcal{Y}+{R_n}^T\mathcal{Y}{R_n}^{-1}+{Q_n}^T\mathcal{Y}{Q_n}^{-1}-{S_n}^T\mathcal{Y}{S_n}^{-1}).$$

Furthermore, we can calculate $\widehat{\mathcal{Y}}$ in terms of $\left\{Y_{ij}\right\}$ and set

$$\widehat{\mathcal{Y}}:=\left[\begin{array}{cccc}{Z}_4& -Z_3& Z_2& Z_1\\ -Z_3& -Z_4& -Z_1& Z_2\\ Z_2& Z_1& Z_4& -Z_3\\ -Z_1& Z_2& -Z_3& -Z_4\end{array}\right],$$

where

$$Z_1=\frac{1}{4}(Y_{14}-Y_{23}+Y_{32}-Y_{41}),Z_2=\frac{1}{4}(Y_{13}+Y_{24}+Y_{31}+Y_{42}),$$

$$Z_3=\frac{1}{4}(-Y_{12}-Y_{21}-Y_{34}-Y_{43}),Z_4=\frac{1}{4}(Y_{11}-Y_{22}+Y_{33}-Y_{44}).$$

Now, we construct a split quaternion matrix X, using the solution $\widehat{\mathcal{Y}}$:

$$X=Z_1+Z_2\mathbf{i}+Z_3\mathbf{j}+Z_4\mathbf{k}=\frac{1}{4}\left[\begin{array}{cccc}{I}_n\mathbf{k}& -I_n\mathbf{j}& I_n\mathbf{i}& -I_n\end{array}\right]\widehat{\mathcal{Y}}\left[\begin{array}{c}{I}_n\\ -I_n\mathbf{i}\\ I_n\mathbf{j}\\ I_n\mathbf{k}\end{array}\right].$$

(13)

Via direct computation, we have $X^{\sigma }=\widehat{\mathcal{Y}}$, and the constructed $X^{\sigma }$ satisfies (9), i.e.,

$${\left({\eta }^{-1}{I}_n\right)}^{\sigma }{G}_n{X}^{\sigma }{G}_n{\left(\eta I_n\right)}^{\sigma }+C^{\sigma }{G}_n{X}^{\sigma }{G}_n{D}^{\sigma }=E^{\sigma }.$$

Using the inverse of ${(·)}^{\sigma }$, we get

$$\left({\eta }^{-1}{I}_n\right)X\left(\eta I_n\right)+CXD=E.$$

Thus, $X^{\eta }+CXD=E$; that is, X in (13) is a solution to (2).

Therefore, (2) is consistent if and only if (9) is consistent. Additionally, we have proven that the solutions to (2) can be constructed from the solution $\mathcal{Y}$ of (9) when (9) is solvable.

To prove (b), suppose that the split quaternion matrix Equation (2) has a solution, $X\in {\mathbb{H}}_s^{n\times n}$, i.e.,

$$X^{\eta \ast }+CXD=E,$$

which is equivalent to

$$X+D^{\eta \ast }{X}^{\eta \ast }{C}^{\eta \ast }=E^{\eta \ast }.$$

Then, using $X^{\eta \ast }={\eta }^{-1}{X}^{\ast }\eta$ produces

$$X+D^{\eta \ast }{\eta }^{-1}{X}^{\ast }\eta C^{\eta \ast }=E^{\eta \ast }.$$

Applying (a) and (c) of Proposition 1 to (2) yields

$$X^{\tau }+{\left(D^{\eta \ast }{\eta }^{-1}\right)}^{\tau }{\left(X^{\ast }\right)}^{\tau }{\left(\eta C^{\eta \ast }\right)}^{\tau }={\left(E^{\eta \ast }\right)}^{\tau },$$

or

$$X^{\tau }+{\left(D^{\eta \ast }{\eta }^{-1}\right)}^{\tau }(-G_n{\left(X^{\tau }\right)}^T{G}_n){\left(\eta C^{\eta \ast }\right)}^{\tau }={\left(E^{\eta \ast }\right)}^{\tau }.$$

Thus, $X^{\tau }$ is a solution to (11).

Conversely, suppose that the real matrix Equation (11) has a solution, $\mathcal{Y}=\left(Y_{ij}\right),Y_{ij}\in {\mathbb{R}}^{n\times n},i,j=1,2,3,4.$ From (b) of Proposition 1, we have

$$\begin{array}{cc}\hfill \mathcal{Y}-\left(R_n^{-1}{\left(D^{\eta \ast }{\eta }^{-1}\right)}^{\tau }{R}_n\right)G_n{\mathcal{Y}}^T{G}_n\left(R_n^{-1}{\left(\eta C^{\eta \ast }\right)}^{\tau }{R}_n\right)& =\left(R_n^{-1}{\left(E^{\eta \ast }\right)}^{\tau }{R}_n\right),\hfill \\ \hfill \mathcal{Y}-\left(Q_n^{-1}{\left(D^{\eta \ast }{\eta }^{-1}\right)}^{\tau }{Q}_n\right)G_n{\mathcal{Y}}^T{G}_n\left(Q_n^{-1}{\left(\eta C^{\eta \ast }\right)}^{\tau }{Q}_n\right)& =\left(Q_n^{-1}{\left(E^{\eta \ast }\right)}^{\tau }{Q}_n\right),\hfill \\ \hfill \mathcal{Y}-\left(S_n^{-1}{\left(D^{\eta \ast }{\eta }^{-1}\right)}^{\tau }{S}_n\right)G_n{\mathcal{Y}}^T{G}_n\left(S_n^{-1}{\left(\eta C^{\eta \ast }\right)}^{\tau }{S}_n\right)& =\left(S_n^{-1}{\left(E^{\eta \ast }\right)}^{\tau }{S}_n\right).\hfill \end{array}$$

(14)

Premultiplying $R_n,Q_n,S_n$, postmultiplying ${R_n}^{-1},{Q_n}^{-1},{S_n}^{-1}$, to both sides of (14) and then using

$$R_n{G}_n=-G_n{R}_n,Q_n{G}_n=G_n{Q}_n,S_n{G}_n=-G_n{S}_n,$$

$$G_n{R_n}^{-1}=-{R_n}^{-1}{G}_n,G_n{Q_n}^{-1}={Q_n}^{-1}{G}_n,G_n{S_n}^{-1}=-{S_n}^{-1}{G}_n$$

to (14) yields

$$\begin{array}{cc}\hfill R_n\mathcal{Y}{R}_n^{-1}-{\left(D^{\eta \ast }{\eta }^{-1}\right)}^{\tau }{G}_n{R}_n{\mathcal{Y}}^T{R_n}^{-1}{G}_n{\left(\eta C^{\eta \ast }\right)}^{\tau }& ={\left(E^{\eta \ast }\right)}^{\tau },\hfill \\ \hfill Q_n\mathcal{Y}{Q}_n^{-1}-{\left(D^{\eta \ast }{\eta }^{-1}\right)}^{\tau }{G}_n{Q}_n{\mathcal{Y}}^T{Q_n}^{-1}{G}_n{\left(\eta C^{\eta \ast }\right)}^{\tau }& ={\left(E^{\eta \ast }\right)}^{\tau },\hfill \\ \hfill S_n\mathcal{Y}{S}_n^{-1}-{\left(D^{\eta \ast }{\eta }^{-1}\right)}^{\tau }{G}_n{S}_n{\mathcal{Y}}^T{S_n}^{-1}{G}_n{\left(\eta C^{\eta \ast }\right)}^{\tau }& ={\left(E^{\eta \ast }\right)}^{\tau }.\hfill \end{array}$$

It is easy to see that $R_n\mathcal{Y}{R_n}^{-1},Q_n\mathcal{Y}{Q_n}^{-1}$ and $S_n\mathcal{Y}{S_n}^{-1}$ are solutions to the real matrix Equation (11), and so is

$$\widehat{\mathcal{Y}}:=\frac{1}{4}(\mathcal{Y}+R_n\mathcal{Y}{R_n}^{-1}+Q_n\mathcal{Y}{Q_n}^{-1}+S_n\mathcal{Y}{S_n}^{-1}).$$

After simplification, we can get $\widehat{\mathcal{Y}}$ in terms of $\left\{Y_{ij}\right\}$ and set

$$\widehat{\mathcal{Y}}:=\left[\begin{array}{cccc}{Z}_1& -Z_2& Z_3& Z_4\\ Z_2& Z_1& Z_4& -Z_3\\ Z_3& Z_4& Z_1& -Z_2\\ Z_4& -Z_3& Z_2& Z_1\end{array}\right],$$

where

$$Z_1=\frac{1}{4}(Y_{11}+Y_{22}+Y_{33}+Y_{44}),Z_2=\frac{1}{4}(-Y_{12}+Y_{21}-Y_{34}+Y_{43}),$$

$$Z_3=\frac{1}{4}(Y_{13}-Y_{24}+Y_{31}-Y_{42}),Z_4=\frac{1}{4}(Y_{14}+Y_{23}+Y_{32}+Y_{41}).$$

Now, we construct a split quaternions matrix from $\widehat{\mathcal{Y}}$ as follows:

$$X=Z_1+Z_2\mathbf{i}+Z_3\mathbf{j}+Z_4\mathbf{k}=\frac{1}{4}\left[\begin{array}{cccc}{I}_n& I_n\mathbf{i}& I_n\mathbf{j}& I_n\mathbf{k}\end{array}\right]\widehat{\mathcal{Y}}\left[\begin{array}{c}{I}_n\\ -I_n\mathbf{i}\\ I_n\mathbf{j}\\ I_n\mathbf{k}\end{array}\right].$$

Via direct computation, we can see that $\widehat{\mathcal{Y}}=X^{\tau }$ is a solution to the real matrix Equation (11). Now, $X^{\tau }$ satisfies

$$X^{\tau }-{\left(D^{\eta \ast }{\eta }^{-1}\right)}^{\tau }{G}_n{\left(X^{\tau }\right)}^T{G}_n{\left(\eta C^{\eta \ast }\right)}^{\tau }={\left(E^{\eta \ast }\right)}^{\tau }.$$

According to (c) of Proposition 1, we get

$$X^{\tau }+{\left(D^{\eta \ast }{\eta }^{-1}\right)}^{\tau }{\left(X^{\ast }\right)}^{\tau }{\left(\eta C^{\eta \ast }\right)}^{\tau }={\left(E^{\eta \ast }\right)}^{\tau },$$

and with the inverse of ${(·)}^{\tau }$,

$$X+D^{\eta \ast }{X}^{\eta \ast }{C}^{\eta \ast }=E^{\eta \ast }.$$

Now, it is easy to see that X is a solution to the split quaternion matrix Equation (2).

In conclusion, (2) is solvable if and only if (11) is solvable. In addition, we have proven that (2) can be constructed using the solution $\mathcal{Y}$ of (11) when (11) is solvable □

Remark 2.

When $★\in \{\ast ,\mathbf{i}\ast ,\mathbf{j}\ast ,\mathbf{k}\ast \}$, Equation (2) was also investigated by [13] via using real representation, a vector operator, and a Kronecker product. In addition, note that the equation $X^{★}-AX{A}^{★}=E$ is the special case of Equation (2), where $★\in \{1,\mathbf{i},\mathbf{j},\mathbf{k},\ast ,\mathbf{i}\ast ,\mathbf{j}\ast ,\mathbf{k}\ast \}$. Thus, according to Theorem 3, we can also derive the corresponding results of the above equation.

5. Numerical Example

In this section, we provide two numerical examples to illustrate our results.

Example 1.

Consider the split quaternion matrix equation $AXB+{\left(AXB\right)}^{\mathbf{i}\ast }=E$, where

$$\begin{array}{cc}\hfill A& =\left[\begin{array}{cc}1& -1\\ 0& 2\end{array}\right]+\left[\begin{array}{cc}-2& 1\\ 4& 3\end{array}\right]\mathbf{i}+\left[\begin{array}{cc}1& 0\\ 1& 2\end{array}\right]\mathbf{j}+\left[\begin{array}{cc}-1& 1\\ 2& 3\end{array}\right]\mathbf{k},\hfill \\ \hfill B& =\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}2& -1\\ 4& 1\end{array}\right]\mathbf{i}+\left[\begin{array}{cc}0& 1\\ 2& -1\end{array}\right]\mathbf{j}+\left[\begin{array}{cc}-1& 2\\ 4& 0\end{array}\right]\mathbf{k},\hfill \\ \hfill E& =\left[\begin{array}{cc}138& -83\\ -83& 42\end{array}\right]+\left[\begin{array}{cc}0& -47\\ 47& 0\end{array}\right]\mathbf{i}+\left[\begin{array}{cc}18& 35\\ 35& -68\end{array}\right]\mathbf{j}+\left[\begin{array}{cc}-122& 43\\ 43& -12\end{array}\right]\mathbf{k}.\hfill \end{array}$$

According to Theorem 2, the corresponding real matrix equation is

$$A^{\tau }Y{B}^{\tau }-{(-\mathbf{i}{B}^{\ast })}^{\tau }{G}_2{Y}^T{G}_2{\left(A^{\ast }\mathbf{i}\right)}^{\tau }=E^{\tau }.$$

(15)

Note that

$$Y=\left[\begin{array}{cccccccc}1& 1& 0& 0& 1& 0& -2& 3\\ -1& 0& -2& -1& 0& 2& 1& -1\\ 0& 0& 1& 1& -2& 3& -1& 0\\ 2& 1& -1& 0& 1& -1& 0& -2\\ 1& 0& -2& 3& 1& 1& 0& 0\\ 0& 2& 1& -1& -1& 0& -2& -1\\ -2& 3& -1& 0& 0& 0& 1& 1\\ 1& -1& 0& -2& 2& 1& -1& 0\end{array}\right]$$

is the solution to the real matrix Equation (15). Thus, according to the Formula (7),

$$X=\left[\begin{array}{cc}1& 1\\ -1& 0\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 2& 1\end{array}\right]\mathbf{i}+\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]\mathbf{j}+\left[\begin{array}{cc}-2& 3\\ 1& -1\end{array}\right]\mathbf{k}$$

is the solution to the split quaternion matrix equation $AXB+{\left(AXB\right)}^{\mathbf{i}\ast }=E$.

Example 2.

Consider the split quaternion matrix equation $X^{\mathbf{j}}+CXD=E$, where

$$\begin{array}{cc}\hfill C& =\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]+\left[\begin{array}{cc}0& 5\\ -1& 3\end{array}\right]\mathbf{i}+\left[\begin{array}{cc}3& -2\\ 0& -1\end{array}\right]\mathbf{j}+\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\mathbf{k},\hfill \\ \hfill D& =\left[\begin{array}{cc}0& 0\\ 2& 0\end{array}\right]+\left[\begin{array}{cc}-3& -2\\ 0& 1\end{array}\right]\mathbf{i}+\left[\begin{array}{cc}4& -1\\ 1& 0\end{array}\right]\mathbf{j}+\left[\begin{array}{cc}0& 1\\ -2& 0\end{array}\right]\mathbf{k},\hfill \\ \hfill E& =\left[\begin{array}{cc}-13& 19\\ 16& 6\end{array}\right]+\left[\begin{array}{cc}-17& 9\\ 7& -2\end{array}\right]\mathbf{i}+\left[\begin{array}{cc}-58& 15\\ -20& -6\end{array}\right]\mathbf{j}+\left[\begin{array}{cc}22& -5\\ 10& 0\end{array}\right]\mathbf{k}.\hfill \end{array}$$

According to Theorem 3, the corresponding real matrix equation is

$${\left(\mathbf{j}{I}_2\right)}^{\sigma }{G}_{2}Y{G}_2{\left(\mathbf{j}{I}_2\right)}^{\sigma }+C^{\sigma }{G}_{2}Y{G}_2{D}^{\sigma }=E^{\sigma }.$$

(16)

Note that

$$Y=\left[\begin{array}{cccccccc}0& 3& 2& -1& 0& 0& 0& 0\\ -1& 2& 0& -1& 0& 0& 1& -2\\ 2& -1& 0& -3& 0& 0& 0& 0\\ 0& -1& 1& -2& -1& 2& 0& 0\\ 0& 0& 0& 0& 0& 3& 2& -1\\ 0& 0& 1& -2& -1& 2& 0& -1\\ 0& 0& 0& 0& 2& -1& 0& -3\\ -1& 2& 0& 0& 0& -1& 1& -2\end{array}\right]$$

is the solution to the real matrix Equation (16). Thus, according to Formula (10),

$$X=\left[\begin{array}{cc}0& 0\\ 1& -2\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\mathbf{i}+\left[\begin{array}{cc}-2& 1\\ 0& 1\end{array}\right]\mathbf{j}+\left[\begin{array}{cc}0& 3\\ -1& 2\end{array}\right]\mathbf{k}$$

is the solution to the split quaternion matrix equation $X^{\mathbf{j}}+CXD=E$.

6. Conclusions

In this paper, we have proposed two real representations of a split quaternion matrix and used them to convert the matrix equation over split quaternions into the problem of a matrix equation over a real number field; there are several kinds of methods available to reach the solutions to those real matrix equations. See, for example, [19,20]. Next, we derived the existence and expression of solutions to the equations $AXB+{\left(AXB\right)}^{★}=E$ and $X^{★}+CXD=E$, where $★\in \{1,\mathbf{i},\mathbf{j},\mathbf{k},\ast ,\mathbf{i}\ast ,\mathbf{j}\ast ,\mathbf{k}\ast \}$, including the special case of $AX+{\left(AX\right)}^{★}=E$ and $X^{★}+AX{A}^{★}=E$. Moreover, based on those results, we have provided two numerical examples to show their feasibility.